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Hausdorff Space

  1. Jan 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that X is Hausdorff if and only if the diagonal [tex]\Delta = \{x \times x | x \in X \}[/tex] is closed in [tex]X \times X[/tex].

    2. Relevant equations

    Definition of Hausdorff Space (T2) : A topological space in which distinct points have disjoint neighborhoods.

    3. The attempt at a solution
     
  2. jcsd
  3. Jan 24, 2009 #2
    With so little work on your part shown to go by it's difficult to know where you're stuck.

    If you haven't already, express the closedness of [tex]\Delta[/tex] in [tex]X\times X[/tex] (which I assume has the product topology) in terms of the openness of its complement.

    Now have a go at proving each direction (neither is more difficult than the other) if you haven't already, and please show us your efforts.
     
  4. Jan 24, 2009 #3
    ->
    If X is Hausdorff, the diagonal [tex] \Delta [/tex] is closed in [tex]X\times X[/tex].

    Assume X is Hausdorff. Now, we have two distinct points x, y and disjoint open sets U, V containing x, y, respectively. The basis element [tex]U \times V[/tex] containing [tex](x,y) \in X \times X[/tex] should not intersect [tex] \Delta [/tex] by the assumption given to the Hausdorff property.
    For every [tex] (x,y) \notin \Delta [/tex], we have a basis element in [tex] X \times X [/tex] containing (x,y), which does not intersect [tex] \Delta [/tex].
    Thus, [tex]X \times X \setminus \Delta is open[/tex] and we conclude [tex] \Delta [/tex] is a closed set in [tex]X \times X[/tex] .

    <-
    If the diagonal [tex] \Delta [/tex] is closed in [tex]X\times X[/tex], X is Haudorff.

    Supppose [tex] \Delta [/tex] is closed in [tex]X\times X[/tex]. Then, [tex]X \times X \setminus \Delta [/tex] is open. Let [tex](x,y) \in X \times X[/tex] and [tex] x \neq y [/tex]. For [tex](x,y) \notin \Delta[/tex], we have a basis element [tex] U \times V [/tex] in [tex]X \times X [/tex] containing (x, y).
    We remain to show U and V are disjoint. Suppose on the contrary that U and V are not disjoint. Then, there is an element [tex] (z,z) \in X times X [/tex] which belongs to both U and V. Contradicting the fact that x and y are distinct.
    Thus, X is Hausdorff.
     
  5. Jan 24, 2009 #4
    That's quite a leap from your previous post.

    Now how 'bout you tackle your other question without copying down the solution from an external source. It's the only way you'll learn topology.
     
  6. Jan 24, 2009 #5
    Can you please give me a link of an external source you mentioned?
    It's my self study of topology (I am not even majoring in math) and I don't need to copy the external source to show you something to impress you. Rather, I just did for my self study purpose and asked an advice if someone finds an error in my attempt to the solution.
     
  7. Jan 24, 2009 #6
    Well, in that case, your work is quite error-free indeed! I would only suggest rephrasing picking an element [tex](x,y)\not\in \Delta[/tex] as picking an element [tex](x,y)\in (X\times X)\backslash \Delta[/tex]. Now, on to your other problem!
     
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